# Cube Numbers as Sum of Sequence of Cubes

## Theorem

 $\ds 16 \, 830^3$ $=$ $\ds \sum_{k \mathop = 1134}^{2133} k^3$ $\ds$ $=$ $\ds 1134^3 + 1135^3 + \cdots + 2133^3$

## Proof

 $\ds \sum_{k \mathop = 1134}^{2133} k^3$ $=$ $\ds \sum_{k \mathop = 1}^{2133} k^3 - \sum_{k \mathop = 1}^{1133} k^3$ $\ds$ $=$ $\ds \frac {2133^2 \paren {2133 + 1}^2} 4 - \frac {1133^2 \paren {1133 + 1}^2} 4$ Sum of Sequence of Cubes $\ds$ $=$ $\ds \frac {\paren {3^3 \times 79}^2 \paren {2 \times 11 \times 97}^2} 4 - \frac {\paren {11 \times 103}^2 \paren {2 \times 3^4 \times 7}^2} 4$ extracting Prime Decomposition $\ds$ $=$ $\ds 3^6 \times 11^2 \paren {\paren {79 \times 97}^2 - \paren {103 \times 3 \times 7}^2}$ factorising $\ds$ $=$ $\ds 3^6 \times 11^2 \paren {7663^2 - 2163^2}$ factorising $\ds$ $=$ $\ds 3^6 \times 11^2 \paren {7663 + 2163} \times \paren {7663 - 2163}$ Difference of Two Squares $\ds$ $=$ $\ds 3^6 \times 11^2 \times 9826 \times 5500$ $\ds$ $=$ $\ds 3^6 \times 11^2 \times \paren {2 \times 17^3} \times \paren {2^2 \times 5^3 \times 11}$ $\ds$ $=$ $\ds 2^3 \times 3^6 \times 5^3 \times 11^3 \times 17^3$ $\ds$ $=$ $\ds \paren {2 \times 3^2 \times 5 \times 11 \times 17}^3$ $\ds$ $=$ $\ds 16 \, 830^3$

$\blacksquare$